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1983AIME 真题及答案解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月12日下午12:37

1983AIME 真题及答案解析

答案解析请参考文末

Problem 1

Let $x$ , $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $log_xw=24$ , $log_y w = 40$ and $log_{xyz}w=12$ . Find $log_zw$ .

Problem 2

Let $f(x)=|x-p|+|x-15|+|x-p-15|$ , where $0 < p < 15$ . Determine the minimum value taken by $f(x)$ for $x$ in the interval $p leq xleq15$ .

Problem 3

What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 sqrt{x^2 + 18x + 45}$ ?

Problem 4

A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.

Pdfresizer.com-pdf-convert-aimeq4.png

Problem 5

Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

Problem 6

Let $a_n=6^{n}+8^{n}$ . Determine the remainder on dividing $a_{83}$ by $49$ .

Problem 7

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Problem 8

What is the largest $2$ -digit prime factor of the integer $n = {200choose 100}$ ?

Problem 9

Find the minimum value of $frac{9x^2sin^2 x + 4}{xsin x}$ for $0 < x < pi$ .

Problem 10

The numbers $1447$ , $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

Problem 11

The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6sqrt{2}$ , what is the volume of the solid?

$[asy] import three; size(170); pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); label("A",A, S); label("B",B, S); label("C",C, S); label("D",D, S); label("E",E,N); label("F",F,N); [/asy]$

Problem 12

Diameter $AB$ of a circle has length a $2$ -digit integer (base ten). Reversing the digits gives the length of the perpendicular chord $CD$ . The distance from their intersection point $H$ to the center $O$ is a positive rational number. Determine the length of $AB$ .

Pdfresizer.com-pdf-convert-aimeq12.png

Problem 13

For ${1, 2, 3, ldots, n}$ and each of its nonempty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract succesive numbers. For example, the alternating sum for ${1, 2, 3, 6,9}$ is $9-6+3-2+1=5$ and for ${5}$ it is simply $5$ . Find the sum of all such alternating sums for $n=7$ .

Problem 14

In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$ .

$[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]$

Problem 15

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $frac{m}{n}$ in lowest terms, what is the product $mn$ ? $[asy]size(140); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]$

1983AIME 详细解析

The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms. $x^{24}=w$ , $y^{40}=w$ , and $(xyz)^{12}=w$ . If we now convert everything to a power of $120$ , it will be easy to isolate $z$ and $w$ . $x^{120}=w^5$ , $y^{120}=w^3$ , and $(xyz)^{120}=w^{10}$ .With some substitution, we get $w^5w^3z^{120}=w^{10}$ and $log_zw=boxed{060}$ .
It is best to get rid of the absolute values first.Under the given circumstances, we notice that $|x-p|=x-p$ , $|x-15|=15-x$ , and $|x-p-15|=15+p-x$ .Adding these together, we find that the sum is equal to $30-x$ , which attains its minimum value (on the given interval $p leq x leq 15$ ) when $x=boxed{015}$ .
If we were to expand by squaring, we would get a quartic polynomial, which isn't always the easiest thing to deal with.Instead, we substitute $y$ for $x^2+18x+30$ , so that the equation becomes $y=2sqrt{y+15}$ .Now we can square; solving for $y$ , we get $y=10$ or $y=-6$ . The second root is extraneous since $2sqrt{y+15}$ is always non-negative (and moreover, plugging in $y=-6$ , we get $-6=6$ , which is obviously false). Hence we have $y=10$ as the only solution for $y$ . Substituting $x^2+18x+30$ back in for $y$ , $x^2+18x+30=10 Longrightarrow x^2+18x+20=0.$ Both of the roots of this equation are real, since its discriminant is $18^2 - 4 cdot 1 cdot 20 = 244$ , which is positive. Thus by Vieta's formulas, the product of the real roots is simply $boxed{020}$ .We begin by noticing that the polynomial on the left is $15$ less than the polynomial under the radical sign. Thus: $[(x^2+ 18x + 45) - 2sqrt{x^2+18x+45} - 15 = 0.]$ Letting $n = sqrt{x^2+18x+45}$ , we have $n^2-2n-15 = 0 Longrightarrow (n-5)(n+3) = 0$ . Because the square root of a real number can't be negative, the only possible $n$ is $5$ .Substituting that in, we have $[sqrt{x^2+18x+45} = 5 Longrightarrow x^2 + 18x + 45 = 25 Longrightarrow x^2+18x+20=0.]$
Reasoning as in Solution 1, the product of the roots is $boxed{020}$ .

Begin by completing the square on both sides of the equation, which gives $[(x+9)^2-51=2sqrt{(x+3)(x+15)}]$ Now by substituting $y=x+9$ , we get $y^2-51=2sqrt{(y-6)(y+6)}$ , or $[y^4-106y^2+2745=0]$ The solutions in $y$ are then $[y=x+9=pm3sqrt{5},pmsqrt{61}]$ Turns out, $pm3sqrt{5}$ are a pair of extraneous solutions. Thus, our answer is then $[left(sqrt{61}-9right)left(-sqrt{61}-9right)=81-61=boxed{020}]$ By difference of squares.
Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Let the foot of the perpendicular from $O$ to $AB$ be $D$ and let the foot of the perpendicular from $O$ to the line $BC$ be $E$ . Let $OE=x$ and $OD=y$ . We're trying to find $x^2+y^2$ . $[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C; pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(D--O--F--B,dashed); dot(O); dot(A); dot(B); dot(C); label("$O$",O,SW); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NE); label("$E$",F,SW); [/asy]$ Applying the Pythagorean Theorem, $OA^2 = OD^2 + AD^2$ and $OC^2 = EC^2 + EO^2$ .Thus, $left(sqrt{50}right)^2 = y^2 + (6-x)^2$ , and $left(sqrt{50}right)^2 = x^2 + (y+2)^2$ . We solve this system to get $x = 1$ and $y = 5$ , such that the answer is $1^2 + 5^2 = boxed{026}$ .Drop perpendiculars from $O$ to $AB$ (with foot $T_1$ ), $M$ to $OT_1$ (with foot $T_2$ ), and $M$ to $AB$ (with foot $T_3$ ). Also, mark the midpoint $M$ of $AC$ .Then the problem is trivialized. Why? $[asy] size(200); pair dl(string name, pair loc, pair offset) { dot(loc); label(name,loc,offset); return loc; }; pair a[] = {(0,0),(0,5),(1,5),(1,7),(-2,6),(-5,5),(-2,5),(-2,6),(0,6)}; string n[] = {"O","$T_1$","B","C","M","A","$T_3$","M","$T_2$"}; for(int i=0;i<a.length;++i) { dl(n[i],a[i],dir(degrees(a[i],false) ) ); draw(a[(i-1)%a.length]--a[i]); }; dot(a); draw(a[5]--a[1]); draw(a[0]--a[3]); draw(a[0]--a[4]); draw(a[0]--a[2]); draw(a[0]--a[5]); draw(a[5]--a[2]--a[3]--cycle,blue+linewidth(0.7)); draw(a[0]--a[8]--a[7]--cycle,blue+linewidth(0.7)); [/asy]$ First notice that by computation, $OAC$ is a $sqrt {50} - sqrt {40} - sqrt {50}$ isosceles triangle, so $AC = MO$ . Then, notice that $angle MOT_2 = angle T_3MO = angle BAC$ . Therefore, the two blue triangles are congruent, from which we deduce $MT_2 = 2$ and $OT_2 = 6$ . As $T_3B = 3$ and $MT_3 = 1$ , we subtract and get $OT_1 = 5,T_1B = 1$ . Then the Pythagorean Theorem tells us that $OB^2 = boxed{026}$ .
One way to solve this problem is by substitution. We have $x^2+y^2=(x+y)^2-2xy=7$ and $x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10$ Hence observe that we can write $w=x+y$ and $z=xy$ .This reduces the equations to $w^2-2z=7$ and $w(7-z)=10$ .Because we want the largest possible $w$ , let's find an expression for $z$ in terms of $w$ . $w^2-7=2z implies z=frac{w^2-7}{2}$ .
Substituting, $w^3-21w+20=0$ , which factorises as $(w-1)(w+5)(w-4)=0$ (the Rational Root Theorem may be used here, along with synthetic division).

The largest possible solution is therefore $x+y=w=boxed{004}$ .

An alternate way to solve this is to let $x=a+bi$ and $y=c+di$ .

Because we are looking for a value of $x+y$ that is real, we know that $d=-b$ , and thus $y=c-bi$ .

Expanding $x^2+y^2=7+0i$ will give two equations, since the real and imaginary parts must match up.

$(a+bi)^2+(c-bi)^2=7+0i$

$(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i$

Looking at the imaginary part of that equation, $2ab-2cb=0$ , so $a=c$ , and $x$ and $y$ are actually complex conjugates.

Looking at the real part of the equation and plugging in $a=c$ , $2a^2-2b^2=7$ , or $2b^2=2a^2-7$ .

Now, evaluating the real part of $(a+bi)^3+(a-bi)^3$ , which equals $10$ (ignoring the odd powers of $i$ , since they would not result in something in the form of $10+0i$ ):

$a^3+3a(bi)^2+a^3+3a(-bi)^2=10$

$2a^3-6ab^2=10$

Since we know that $2b^2=2a^2-7$ , it can be plugged in for $b^2$ in the above equation to yield:

$2a^3-3a(2a^2-7)=10$

$-4a^3+21a=10$

$4a^3-21a+10=0$

Since the problem is looking for $x+y=2a$ to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, $a=10, a=5, a=frac{5}{2}$ all fail, but $a=2$ does work. Thus, the real part of both numbers is $2$ , and their sum is $boxed{004}$ .

Begin by assuming that $x$ and $y$ are roots of some polynomial of the form $w^2+bw+c$ , such that by Vieta's Formulae and some algebra (left as an exercise to the reader), $b^2-2c=7$ and $3bc-b^3=10$ . Substituting $c=frac{b^2-7}{2}$ , we deduce that $b^3-21b-20=0$ , whose roots are $-4$ , $-1$ , and $5$ . Since $-b$ is the sum of the roots and is maximized when $b=-4$ , the answer is $-(-4)=boxed{004}$ .
Firstly, we try to find a relationship between the numbers we're provided with and $49$ . We notice that $49=7^2$ , and both $6$ and $8$ are greater or less than $7$ by $1$ .Thus, expressing the numbers in terms of $7$ , we get $a_{83} = (7-1)^{83}+(7+1)^{83}$ .Applying the Binomial Theorem, half of our terms cancel out and we are left with $2(7^{83}+3403cdot7^{81}+cdots + 83cdot7)$ . We realize that all of these terms are divisible by $49$ except the final term.After some quick division, our answer is $boxed{035}$ .
We can use complementary counting, by finding the probability that none of the three knights are sitting next to each other and subtracting it from $1$ .Imagine that the $22$ other (indistinguishable) people are already seated, and fixed into place.We will place $A$ , $B$ , and $C$ with and without the restriction.There are $22$ places to put $A$ , followed by $21$ places to put $B$ , and $20$ places to put $C$ after $A$ and $B$ . Hence, there are $22cdot21cdot20$ ways to place $A, B, C$ in between these people with restrictions.Without restrictions, there are $22$ places to put $A$ , followed by $23$ places to put $B$ , and $24$ places to put $C$ after $A$ and $B$ . Hence, there are $22cdot23cdot24$ ways to place $A,B,C$ in between these people without restrictions.Thus, the desired probability is $1-frac{22cdot21cdot20}{22cdot23cdot24}=1-frac{420}{552}=1-frac{35}{46}=frac{11}{46}$ , and the answer is $11+46=boxed{057}$ .
Expanding the binomial coefficient, we get ${200 choose 100}=frac{200!}{100!100!}$ . Let the required prime be $p$ ; then $10 le p < 100$ . If $p > 50$ , then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$ . The largest such prime is $boxed{061}$ , which is our answer.
Let $y=xsin{x}$ . We can rewrite the expression as $frac{9y^2+4}{y}=9y+frac{4}{y}$ .Since $x>0$ , and $sin{x}>0$ because $0< x<pi$ , we have $y>0$ . So we can apply AM-GM: $[9y+frac{4}{y}ge 2sqrt{9ycdotfrac{4}{y}}=12]$ The equality holds when $9y=frac{4}{y}Longleftrightarrow y^2=frac49Longleftrightarrow y=frac23$ .Therefore, the minimum value is $boxed{012}$ . This is reached when we have $x sin{x} = frac{2}{3}$ in the original equation (since $xsin x$ is continuous and increasing on the interval $0 le x le frac{pi}{2}$ , and its range on that interval is from $0 le xsin x le frac{pi}{2}$ , this value of $frac{2}{3}$ is attainable by the Intermediate Value Theorem).
Suppose that the two identical digits are both $1$ . Since the thousands digit must be $1$ , only one of the other three digits can be $1$ . This means the possible forms for the number are
$11xy,qquad 1x1y,qquad1xy1$

Because the number must have exactly two identical digits, $xneq y$ , $xneq1$ , and $yneq1$ . Hence, there are $3cdot9cdot8=216$ numbers of this form.

Now suppose that the two identical digits are not $1$ . Reasoning similarly to before, we the following possibilities:

$1xxy,qquad1xyx,qquad1yxx.$

Again, $xneq y$ , $xneq 1$ , and $yneq 1$ . There are $3cdot9cdot8=216$ numbers of this form.

Thus the answer is $216+216=boxed{432}$ .

Consider a sequence of $4$ digits instead of a $4$ -digit number. Only looking at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is $frac{1}{10}$ . This means we can find all possible sequences with one digit repeated twice, and then divide by $10$ .

If we let the three distinct digits of the sequence be $a, b,$ and $c$ , with $a$ repeated twice, we can make a table with all possible sequences:

$[begin{tabular}{ccc} aabc & abac & abca \ baac & baca & \ bcaa && \ end{tabular}]$

There are $6$ possible sequences.

Next, we can see how many ways we can pick $a$ , $b$ , and $c$ . This is $10(9)(8) = 720$ , because there are $10$ digits, from which we need to choose $3$ with regard to order. This means there are $720(6) = 4320$ sequences of length $4$ with one digit repeated. We divide by 10 to get $boxed{432}$ as our answer.
First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$ . The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$ , and one of the legs is $3sqrt{2}$ . We apply the Pythagorean Theorem to deduce that the height is $6$ . $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(B--Ba--Ca--C,dashed+d); draw(A--Aa--Da--D,dashed+d); draw(E--(E.x,E.y,0),dashed+l); draw(F--(F.x,F.y,0),dashed+l); draw(Aa--E--Da,dashed+d); draw(Ba--F--Ca,dashed+d); label("A",A,S); label("B",B,S); label("C",C,S); label("D",D,NE); label("E",E,N); label("F",F,N); label("$12sqrt{2}$",(E+F)/2,N); label("$6sqrt{2}$",(A+B)/2,S); label("6",(3*s/2,s/2,3),ENE); [/asy]$ Next, we complete the figure into a triangular prism, and find its volume, which is $frac{6sqrt{2}cdot 12sqrt{2}cdot 6}{2}=432$ .Now, we subtract off the two extra pyramids that we included, whose combined volume is $2cdot left( frac{6sqrt{2}cdot 3sqrt{2} cdot 6}{3} right)=144$ .Thus, our answer is $432-144=boxed{288}$ .
Let $AB=10x+y$ and $CD=10y+x$ . It follows that $CO=frac{AB}{2}=frac{10x+y}{2}$ and $CH=frac{CD}{2}=frac{10y+x}{2}$ . Applying the Pythagorean Theorem on $CO$ and $CH$ , we deduce $[OH=sqrt{left(frac{10x+y}{2}right)^2-left(frac{10y+x}{2}right)^2}=sqrt{frac{9}{4}cdot 11(x+y)(x-y)}=frac{3}{2}sqrt{11(x+y)(x-y)}]$ Because $OH$ is a positive rational number, the quantity $sqrt{11(x+y)(x-y)}$ must be rational, so $11(x+y)(x-y)$ must be a perfect square. Hence either $x-y$ or $x+y$ must be a multiple of $11$ , but as $x$ and $y$ are digits, $1+0=1 leq x+y leq 9+9=18$ , so the only possible multiple of $11$ is $11$ itself. However, $x-y$ cannot be 11, because both must be digits. Therefore, $x+y$ must equal $11$ and $x-y$ must be a perfect square. The only pair $(x,y)$ that satisfies this condition is $(6,5)$ , so our answer is $boxed{065}$ .
Let $S$ be a non- empty subset of ${1,2,3,4,5,6}$ .Then the alternating sum of $S$ , plus the alternating sum of $S cup {7}$ , is $7$ . This is because, since $7$ is the largest element, when we take an alternating sum, each number in $S$ ends up with the opposite sign of each corresponding element of $Scup {7}$ .Because there are $2^{6} - 1 = 63$ of these pairs of sets (subtracting $1$ to exclude the empty set), the sum of all possible subsets of our given set is $63 cdot 7$ . However, we forgot to include the subset that only contains $7$ , so the answer is $64 cdot 7=boxed{448}$ .Note: The empty set is the same as the subset that only includes $7$ so you could have just left it as $2^{6}$ pairs of sets.
Note that some of these solutions assume that $R$ lies on the line connecting the centers, which is not true in general. It is true here only because the perpendicular from $P$ passes through through the point where the line between the centers intersects the small circle. This fact can be derived from the application of the Midpoint Theorem to the trapezoid made by dropping perpendiculars from the centers onto $QR$ .Firstly, notice that if we reflect $R$ over $P$ , we get $Q$ . Since we know that $R$ is on circle $B$ and $Q$ is on circle $A$ , we can reflect circle $B$ over $P$ to get another circle (centered at a new point $C$ , and with radius $6$ ) that intersects circle $A$ at $Q$ . The rest is just finding lengths, as follows.Since $P$ is the midpoint of segment $BC$ , $AP$ is a median of $triangle ABC$ . Because we know $AB=12$ , $BP=PC=6$ , and $AP=8$ , we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = sqrt{56}$ . Now we have a kite $AQCP$ with $AQ=AP=8$ , $CQ=CP=6$ , and $AC=sqrt{56}$ , and all we need is the length of the other diagonal $PQ$ . The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$ . Then $sqrt{36-x^2} + sqrt{64-x^2} = sqrt{56}.$ Solving this equation, we find that $x^2=frac{65}{2}$ , so $PQ^2 = 4x^2 = boxed{130}.$
-Credit to Adamz for diagram- $[asy] size(10cm); import olympiad; pair O = (0,0);dot(O);label("$O$",O,SW); pair M = (4,0);dot(M);label("$M$",M,SE); pair N = (4,2);dot(N);label("$N$",N,NE); draw(circle(O,5)); pair B = (4,3);dot(B);label("$B$",B,NE); pair C = (4,-3);dot(C);label("$C$",C,SE); draw(B--C);draw(O--M); pair P = (1.5,2);dot(P);label("$P$",P,W); draw(circle(P,2.5)); pair A=(3,4);dot(A);label("$A$",A,NE); draw(O--A); draw(O--B); pair Q = (1.5,0); dot(Q); label("$Q$",Q,S); pair R = (3,0); dot(R); label("$R$",R,S); draw(P--Q,dotted); draw(A--R,dotted); pair D=(5,0); dot(D); label("$D$",D,E); draw(A--D); [/asy]$ Let $A$ be any fixed point on circle $O$ , and let $AD$ be a chord of circle $O$ . The locus of midpoints $N$ of the chord $AD$ is a circle $P$ , with diameter $AO$ . Generally, the circle $P$ can intersect the chord $BC$ at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle $P$ is tangent to $BC$ at point $N$ .Let $M$ be the midpoint of the chord $BC$ . From right triangle $OMB$ , we have $OM = sqrt{OB^2 - BM^2} =4$ . This gives $tan angle BOM = frac{BM}{OM} = frac 3 4$ .Notice that the distance $OM$ equals $PN + PO cos angle AOM = r(1 + cos angle AOM)$ , where $r$ is the radius of circle $P$ .Hence $[cos angle AOM = frac{OM}{r} - 1 = frac{2OM}{R} - 1 = frac 8 5 - 1 = frac 3 5]$ (where $R$ represents the radius, $5$ , of the large circle given in the question). Therefore, since $angle AOM$ is clearly acute, we see that $[tan angle AOM =frac{sqrt{1 - cos^2 angle AOM}}{cos angle AOM} = frac{sqrt{5^2 - 3^2}}{3} = frac 4 3]$ Next, notice that $angle AOB = angle AOM - angle BOM$ . We can therefore apply the subtraction formula for $tan$ to obtain $[tan angle AOB =frac{tan angle AOM - tan angle BOM}{1 + tan angle AOM cdot tan angle BOM} =frac{frac 4 3 - frac 3 4}{1 + frac 4 3 cdot frac 3 4} = frac{7}{24}]$ It follows that $sin angle AOB =frac{7}{sqrt{7^2+24^2}} = frac{7}{25}$ , such that the answer is $7 cdot 25=boxed{175}$ .